Question

Two point charges $+2\mu C$ and $+8\mu C$ are placed at a distance of 15 cm apart in air. At a point on the line joining these two charges where the net electric field becomes zero, a third charge of $+5\mu C$ is placed. The net electrostatic force acting on $+5\mu C$ charge is

• A. 16 N
• B. 4 N
• C. 8 N
• D. Zero

Hint:

No calculation of distance is required. If the field is zero, the force on any test charge placed there is zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem states that the third charge is placed at the point where the net electric field is zero (Null Point).
Step 2: Force Calculation:
The electrostatic force $F$ acting on a charge $q$ in an electric field $E$ is given by $F = qE$. Since the charge $+5\mu C$ is placed at the location where the net electric field $E_{net} = 0$: \[ F_{net} = q \times E_{net} = 5\mu C \times 0 = 0 \]
Step 4: Final Answer:
The net force is Zero.